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7/27/2019 Zivot - Lectures on Structural Change.pdf
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Lectures on Structural Change
Eric Zivot
Department of Economics, University of Washington
April 5, 2003
1 Overview of Testing for and Estimating Struc-
tural Change in Econometric Models
1. Day 1: Tests of Parameter Constancy
2. Day 2: Estimation of Models with Structural Change
3. Day 3: Time Varying Parameter Models
2 Some Preliminary Asymptotic Theory
Reference: Stock, J.H. (1994) Unit Roots, Structural Breaks and Trends,in Handbook of Econometrics, Vol. IV.
3 Tests of Parameter Constancy in Linear Mod-
els
3.1 Motivation
Diagnostics for model adequacy
Provide information about out-of-sample forecasting accuracy
Within-sample parameter constancy is a necessary condition for super-exogeneity
3.2 Example Data Sets
3.2.1 Simulated Data
Consider the linear regression model
yt = + xt + t, t = 1, . . . , T = 200
xt iid N(0, 1)t iid N(0, 2)
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No structural change parameterization: = 0, = 1, = 0.5Structural change cases
Break in intercept: = 1 for t > 100
Break in slope: = 3 for t > 100
Break in error variance: = 0.25 for t > 100
Random walk in slope: = t = t1 + t, t iid N(0, 0.1) and 0 = 1.
(show simulated data)
3.2.2 US/DM Monthly Exchange rate data
Let
st = log of spot exchange rate in month t
ft = log of forward exchange rate in month t
The forward rate unbiased hypothesis is typically investigated using the so-calleddifferences regression
st+1 = + (ft st) + t+1ft st = iUSt iDMt = forward discount
If the forward rate ft is an unbiased forecast of the future spot rate st+1 thenwe should find
= 0 and = 1
The forward discount is often modeled as an AR(1) model
ft st = + (ft1 st1) + utStatistical Issues
st+1 is close to random walk with large variance
ft st behaves like highly persistent AR(1) with small variance ft st appears to be unstable over time
3.3 Chow Forecast Test
Reference: Chow, G.C. (1960). Tests of Equality between Sets of Coeffi-cients in Two Linear Regressions, Econometrica, 52, 211-22.
Consider the linear regression model with k variables
yt = x0
t + ut, ut (0, 2), t = 1, . . . , ny = X + u
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Parameter constancy hypothesis
H0 : is constant
Intuition
If parameters are constant then out-of-sample forecasts should be unbiased(forecast errors have mean zero)
Test construction:
Split sample into n1 > k and n2 = n n1 observations
y =
y1y2
n1n2
, X =
X1X2
n1n2
Fit model using first n1
observations
1 = (X0
1X1)1X01y1
u1 = y1 X1121 = u
0
1u1/(n1 k)
Use 1 and X2 to predict y2 using next n2 observations
y2 = X21
Compute out-of-sample prediction errors
u2 = y2y2 = y2X21
Under H0 : is constant
u2 = u2 X2(1)
and
E[u2] = 0
var(u2) = 2
In2 + X2(X0
1X1)1X02
Further, If the errors u are Gaussian then
u2 N(0,var(u2))u02var(u2)
1u2 2(n2)(n1 k)21/2 2(n1 k)
This motivates the Chow forecast test statistic
ChowFCST(n2) =u02
In2 + X2(X
0
1X1)1X02
u2
n221
F(n2, n1 k)
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Decision: Reject H0 at 5% level if
ChowFCST(n2) > cv0.05
Remarks:
Test is a general specification test for unbiased forecasts
Popular with LSE methodology
Implementation requires a priori splitting of data into fit and forecastsamples
3.3.1 Application: Simulated Data
Chow Forecast Test
n2Model 100 50 25No SC 1.121 1.189 1.331Mean shift 9.130*** 1.329* 1.061Slope shift 9.055*** 2.067*** 1.545*Var shift 0.568 0.726 0.864RW slope 2.183*** 1.302 0.550
3.4 CUSUM and CUSUMSQ Tests
Reference: Brown, R.L., J. Durbin and J.M. Evans (1975). Techniquesfor Testing the Constancy of Regression Relationships over Time, Journal ofthe Royal Statistical Society, Series B, 35, 149-192.
3.4.1 Recursive least squares estimation
The recursive least squares (RLS) estimates of are based on estimating
yt = 0
txt + t, t = 1, . . . , n
by least squares recursively for t = k + 1, . . . , n giving nk least squares (RLS)estimates (k+1, . . . , T).
RLS estimates may be efficiently computed using the Kalman Filter
If is constant over time then t should quickly settle down near a com-mon value.
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3.4.2 Recursive residuals
Formal tests for structural stability of the regression coefficients may be com-puted from the standardized 1 step ahead recursive residuals
wt =vt
ft=
yt 0
t1xtft
ft = 2h
1 + x0t(X0
tXt)1xt
iIntuition:
Ifi changes in the next period then the forecast error will not have meanzero
wt are recursive Chow Forecast t-statistics with n2 = 1
3.4.3 CUSUM statistic
The CUSUM statistic of Brown, Durbin and Evans (1975) is
CUSUMt =tX
j=k+1
wjw
2w =1
n knXt=1
(wt w)2
Under the null hypothesis that is constant, CUSUMt has mean zero andvariance that is proportional to t
k
1.
3.4.4 CUSUMSQ statistic
THE CUSUMSQ statistic is
CUSUMSQt =
Ptj=k+1 w
2jPn
j=k+1 w2j
Under the null that is constant, CUSUMSQt behaves like a 2(t) and con-
fidence bounds can be easily derived.
3.4.5 Application: Simulated Data
(insert graphs here)
Remarks
Ploberger and Kramer (1990) show the CUSUM test can be constructedwith OLS residuals instead of recursive residuals
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CUSUM Test is essentially a test to detect instability in intercept alone
CUSUM Test has power only in direction of the mean regressors
CUSUMSQ has power for changing variance
There are tests with better power
3.4.6 Application: Exchange Rate Regression
(insert graphs here)
3.5 Nybloms Parameter Stability Test
Reference:Nyblom, J. (1989). Testing for the Constancy of Parameters OverTime, Journal of the American Statistical Association, 84 (405), 223-230.
Consider the linear regression model with k variables
yt = x0
t + t, t = 1, . . . , n
The time varying parameter (TVP) alternative model assumes
= t = t1 + t, it (0, 2i
), i = 1, . . . , k
The hypotheses of interest are
H0 : is constant 2i
= 0 for all i
H1 : 2i
> 0 for some i
Nyblom (1989) derives the locally best invariant test as the Lagrange multiplier
test. The score assuming Gaussian errors isnXt=1
xtt = 0
t = yt x0t= (X0X)1X0y
Define
ft = xtt
St =t
Xj=1ft = cumulative sums
V = n1X0X
Note thatnXj=1
ft = 0
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Nyblom derives the LM statistic
L = 1n2
nXt=1
StV1St
=1
n2tr
"V1
nXt=1
StS0
t
#
Under mild assumptions regarding the behavior of the regressors, the limitingdistribution of L under the null is a Camer-von Mises distribution:
L Z10
Bk()Bk()
0d
Bk
() = Wk() Wk(1)W
k() = k dimensional Brownian motion
Decision: Reject H0 at 5% level if
L > cv0.05
Remarks:
Distribution of L is non-standard and depends on k.
Critical values are computed by simulation and are given in Nyblom,Hansen (1992) and Hansen (1997)
Test is for constancy of all parameters
Test is not informative about the date or type of structural change
Test is applicable for models estimated by methods other than OLS
Distribution of L is different if xt is non-stationary (unit root, determin-istic trend). See Hansen (1992).
3.5.1 Application: Simulated Data
Nyblom TestModel LcNo SC .332Mean shift 13.14
Slope shift 14.13
var shift .351RW slope 9.77
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3.5.2 Application: Exchange rate regression
Nyblom TestModel LcAR(1) 1.27
Diff reg .413
3.6 Hansens Parameter Stability Tests
References
1. Hansen, B.E. (1992). Testing for Parameter Instability in Linear Mod-els Journal of Policy Modeling, 14(4), 517-533.
2. Hansen, B.E. (1992). Tests for Parameter Instability in Regressionswith I(1) Processes, Journal of Business and Economic Statistics, 10,321-336.
Idea: Extension of Nybloms LM test to individual coefficients.Under the null of constant parameters, the score vector from the linear model
with Gaussian errors is
nXt=1
xitt = 0, i = 1, . . . , kX(2t 2) = 0
t = yt x0t
2 = n1n
Xt=1 2tDefine
fit =
xitt
2t 2i = 1, . . . ki = k + 1
Sit =tX
j=1
fij , i = 1, . . . , k + 1
Note thatn
Xi=1fit = 0, i = 1, . . . , k + 1
3.6.1 Individual Coefficient Tests
Hansens LM test for
H0 : i is constant, i = 1, . . . , k
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and forH0 :
2 is constant
is
Li =1
nVi
nXt=1
S2it, i = 1, . . . , k
Vi =nXt=1
f2it
Under H0 : i is constant or H0 : 2 is constant
Li Z10
B1 ()B1 ()d
Decision: Reject H0 at 5% level if
Li > cv0.05 = 0.470
3.6.2 Joint Test for All Coefficients
For testing the joint hypothesis
H0 : and 2 are constant
define the (k + 1) 1 vectors
ft = (f1t, . . . , f k+1,t)0
St = (S1t, . . . , S k+1,t)0
Hansens LM statistic for testing the constancy of all parameters is
Lc =1
n
nXt=1
S0tV1St =
1
ntr
V1
nXt=1
StS0
t
!
V =nXt=1
ftf0
t
Under the null of no-structural change
Lc
Z1
0
Bk+1()Bk+1()d
Decision: Reject H0 at 5% level if
Lc > cv0.05
Remarks
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Tests are very easy to compute and are robust to heteroskedasticity
Null distribution is non-standard and depends upon number of parameterstested for stability
Individual tests are informative about the type of structural change
Tests are not informative about the date of structural change
Hansens L1 test for constancy of intercept is analogous to the CUSUMtest
Hansens Lk+1 test for constancy of variance is analogous to CUSUMSQtest
Hansens Lc test for constancy of all parameters is similar to Nybolomstest
Distribution of tests is different if data are nonstationary (unit root, de-terministic trend) - see Hansen (1992), JBES.
3.6.3 Application: Simulated Data
Hansen TestsModel 2 JointNo SC .179 .134 .248 .503Mean shift 13.19 .234 .064 13.3
Slope shift .588 5.11 .067 5.25
var shift .226 .119 .376 .736RW slope .253 4.08 .196 4.4
3.6.4 Application: Exchange rate regression contd
Hansen TestsModel intercept slope variance JointAR(1) .382 .147 2.94 3.90
Diff reg .104 .153 .186 .520
4 Tests for Single Structural Change
Consider the linear regression model with k variables
yt = x0
tt + t, t = 1 . . . , n
No structural change null hypothesis
H0 : t=
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Single break date alternative hypothesis
H1 : t= , t m = break datet = + , t > m and 6= 0k < m < n k =
m
n= break fraction
Remarks:
Under no break null = 0.
Pure structural change model: all coefficients change (i 6= 0 for i =1, . . . , k)
Partial structural change model: some coefficients change (i 6= 0 for somei)
m = [ n], [] = integer part
4.1 Chows Test with Known Break Date
Assume: m or is knownFor a data interval [r , . . . , s] such that s r > k define
r,s = OLS estimate of
r,s = OLS residual vector
SS Rr,s = 0
r,sr,s = sum of squared residuals
Chows breakpoint test for testing H0 vs. H1 with m known is
Fn
mn
= Fn () =
(SS R1,n (SS R1,m + SS Rm+1,n))/k(SS R1,m + SS Rm+1,n) /(n 2k)
The Chow test may also be computed as the F-statistic for testing = 0 fromthe dummy variable regression
yt = x0
t + Dt(m)x0
t+ t
Dt(m) = 1 if t > m; 0 otherwise
Under H0 : = 0 with m known
Fn() F(k, n 2k)
k Fn()d
2(k)Decision: Reject H0 at 5% level if
Fn() > F0.95(k, n k)k Fn() >
20.95(k)
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4.1.1 Application: Simulated Data
Chow Breakpoint TestF200(0.5) F200(0.25) F200(0.75)No SC 0.808 0.081 1.55Mean shift 377*** 13.03*** 11.21***Slope shift 374*** 10.97*** 17.57***Var shift 1.071 0.117 1.204RW slope 80.14*** 4.058** 2.218
4.2 Quandts LR Test with Unknown Break Date
References:
1. Quandt, R.E. (1960). Tests of Hypotheses that a Linear System Obeys
Two Separate Regimes, Journal of the American Statistical Association,55, 324-330.
2. Davies, R.A. (1977). Hypothesis Testing When a Nuisance Parameteris Present only Under the Alternative, Biometrika, 64, 247-254.
3. Kim, H.-J., and D. Siegmund (1989). The Likelihood Ratio Test fora Change-Point in Simple Linear Regression, Biometrika, 76, 3, 409-23.
4. Andrews, D.W.K. (1993). Tests for Parameter Instability and Struc-tural Change with Unknown Change Point, Econometrica, 59, 817-858.
5. Hansen, B.E. (1997). Approximate Asymptotic P Values for Structural-Change Tests, Journal of Business and Economic Statistics, 15, 60-67.
Assume: m or is unknown.Quandt considered the LR statistic for testing H0 : = 0 vs. H1 : 6= 0
when m is unknown. This turns out to be the maximal Fn() statistic over arange of break dates m0, . . . , m1 :
QLR = maxm[m0,m1]
Fn
mn
= max
[0,1]Fn()
i =min
= trimming parameters, i = 0, 1
Remarks
QLR is also know as Andrews supF statistic
Trimming parameters 0 and 1 must be set
Cannot have 0 = 1 and 1 = 1 because breaks are hard to identifynear beginning and end of sample
Information about location of break can be used to specify 0 and1
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Andrews recommends 0 = 0.15 and 1 = 0.85 if there is no knowl-edge of break date
Implicitly, the break data m and break fraction are estimated using
m = arg maxm
Fn
mn
= m/n
Under the null, m defined under the alternative is not identified. This isan example of the Davies problem.
Davies (1977) showed that if estimated parameters are unidentified underthe null, standard 2 inference does not obtain.
Under H0 : = 0, Kim and Siegmund (1989) showed
k QLR sup[0,1]
Bk()0Bk()
(1 )Bk() = Wk() Wk(1) = Brownian Bridge
Decision: Reject H0 at 5% level if
k QLR > cv0.05
Remarks
Distribution of QLR is non-standard and depends on the number of vari-ables k and the trimming parameters 0 and 1
Critical values for various values of 0 and 1 computed by simulation aregiven in Andrews (1993), and are larger than 2(k) critical values. For0 = 0.15 and 1 = 0.85
5% critical valuesk 2(k) QLR k QLR1 3.84 8.8510 18.3 27.03
P-values can be computed using techniques from Hansen (1997)
Graphical plot of Fn() statistics is informative to locate the break date
4.2.1 Application: Simulated data
QLR or sup-F Test
QLR
bm
b
No SC 2.87 142 0.71Mean shift 377*** 101 0.51Slope shift 374*** 101 0.51Var shift 2.36 142 0.71RW slope 113*** 77 0.39
(insert graphs here)
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4.2.2 Application: Exchange rate data
QLR or sup-F TestQLR bm b
AR(1) 12.13*** 1989:05 0.65Diff reg 4.08 1991:03 0.74
4.3 Optimal Tests with Unknown Break Date
References:
1. Andrews, D.W.K. and W. Ploberger (1994). Optimal Tests Whena Nuisance Parameter Is Present Only Under the Alternative, Economet-rica, 62, 1383-1414.
Andrews and Ploberger (1994) derive tests for structural change with anunknown break date with optimal power. These tests turn out to be weightedaverages of the Chow breakpoint statistics Fn(
mn
) used to compute the QLRstatistic:
ExpFn = ln
1
m2 m1 + 1m2X
t=m1
exp
1
2k Fn
t
n
!
AveFn =1
m2 m1 + 1m2Xt=m1
k Fn
t
n
k = number of regressors being tested
Remarks
Asymptotic null distributions are non-standard and depend on k, 0 and1
Critical values are given in Andrews and Ploberger; P-values can be com-puted using techniques of Hansen (1997)
Tests can have higher power than QLR statistic
Tests are not informative about location of break date
4.4 Empirical Application
Reference: Stock and Watson (199?), Journal of Business and Economic
Statistics.
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